Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $q = \dfrac{y + 2}{9y^2 - 9y} \times \dfrac{y^2 + 3y - 4}{6y + 12} $
Solution: First factor the quadratic. $q = \dfrac{y + 2}{9y^2 - 9y} \times \dfrac{(y - 1)(y + 4)}{6y + 12} $ Then factor out any other terms. $q = \dfrac{y + 2}{9y(y - 1)} \times \dfrac{(y - 1)(y + 4)}{6(y + 2)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (y + 2) \times (y - 1)(y + 4) } { 9y(y - 1) \times 6(y + 2) } $ $q = \dfrac{ (y + 2)(y - 1)(y + 4)}{ 54y(y - 1)(y + 2)} $ Notice that $(y + 2)$ and $(y - 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ \cancel{(y + 2)}(y - 1)(y + 4)}{ 54y\cancel{(y - 1)}(y + 2)} $ We are dividing by $y - 1$ , so $y - 1 \neq 0$ Therefore, $y \neq 1$ $q = \dfrac{ \cancel{(y + 2)}\cancel{(y - 1)}(y + 4)}{ 54y\cancel{(y - 1)}\cancel{(y + 2)}} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $q = \dfrac{y + 4}{54y} ; \space y \neq 1 ; \space y \neq -2 $